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Many thanks !
For 2 pairings with Canada vs Canada, is it correct to write:
P(C1,C2,CM3,...,CM10) = (11/20*10/19)*(9/18*8/17)*(7/16*7/15*2)*...*(2/4*2/3*2)*(1/2*1/1*2) ?
And the number of arrangements of this is combin(10,2)=45 ?
It seemed to me that COMBIN(11,i)/COMBIN(20,i) is the probability to end up with i pairing from Canada...
But this doesn't seem to work.
Many thanks Bob for your reply.
I understand your method, even though I'm not sure we have to divide par 2 to account for the same pairings (e.g. C1C2 and C2C1)...
I tried to generalize your method for the 20 teams, but I end up with a huge number of possible combinations of only two Cs drawn together.
So I don't think this is working..
Hi,
I'm stuck with this football draw problem: could you help me out ?
I have 20 football teams: 11 are from Canada, 9 are from Mexico.
I will draw randomly 10 pairings out of those 20 teams so as to create 10 matches...
What are the probabilities to have :
- only 1 pairing comprised of 2 canadian teams ?
- only 1 pairing comprised of 2 mexican teams ?
- exactly 2 pairings each comprised of 2 canadian teams ?
- exactly 2 pairings each comprised of 2 mexican teams ?
- exactly 3 pairings each comprised of 2 canadian teams ?
- exactly 3 pairings each comprised of 2 mexican teams ?
- exactly 4 pairings each comprised of 2 canadian teams ?
- exactly 4 pairings each comprised of 2 mexican teams ?
- at least 4 pairings each comprised of 2 canadian teams ?
- at least 4 pairings each comprised of 2 mexican teams ?
Many thanks for your help !
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